Đặt \(A=\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+...+\dfrac{1}{1985}\)

\(A=\dfrac{1}{5}.\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{397}\right)\)

\(A=\dfrac{1}{5}.\left(1+\dfrac{1}{1+2}+\dfrac{1}{2+3}+...+\dfrac{1}{198+199}\right)\)

\(A=\dfrac{1}{5}.\left(1+1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{198}-\dfrac{1}{199}\right)\)

\(A=\dfrac{1}{5}.\left(2-\dfrac{1}{199}\right)\)

\(A=\dfrac{397}{995}< \dfrac{9}{20}\)

\(\Rightarrow\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+...+\dfrac{1}{1985}< \dfrac{9}{20}\left(đpcm\right)\)

Ta có :

\(A=\dfrac{1}{5}+\dfrac{1}{15}+\dfrac{1}{25}+\dfrac{1}{35}+...+\dfrac{1}{1985}\)

\(A=\dfrac{1}{5}+\dfrac{1}{3.5}+\dfrac{1}{5.5}+\dfrac{1}{7.5}+...+\dfrac{1}{397.5}\)

\(\Rightarrow5A=1+\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{397}\)

\(5A-1=\dfrac{1}{3}+\dfrac{1}{5}+\dfrac{1}{7}+...+\dfrac{1}{397}\)

\(5A-1=\dfrac{1}{3}+\left(\dfrac{1}{5}+\dfrac{1}{7}+\dfrac{1}{9}\right)+\left(\dfrac{1}{11}+\dfrac{1}{13}+...+\dfrac{1}{27}\right)+\)

\(\left(\dfrac{1}{29}+\dfrac{1}{31}+...+\dfrac{1}{81}\right)+\left(\dfrac{1}{83}+\dfrac{1}{85}+...+\dfrac{1}{243}\right)+...+\dfrac{1}{397}\)

\(\Rightarrow5A-1>\dfrac{1}{3}+\dfrac{1}{9}.3+\dfrac{1}{27}.9+\dfrac{1}{81}.27+\dfrac{1}{243}.81=\dfrac{1}{3}.5=\dfrac{5}{3}\)

\(\Rightarrow5A-1>\dfrac{5}{4}\Rightarrow5A>\dfrac{9}{4}\)

\(\Rightarrow A>\dfrac{9}{4}:5=\dfrac{9}{20}\Rightarrow\left(dpcm\right)\)